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\begin{center}
\vskip 1cm{\LARGE\bf On Intervals $(kn,(k+1)n)$ Containing a Prime \\
\vskip .1in
for All $n>1$}
\vskip 1cm
\normalsize
Vladimir Shevelev\\
Department of Mathematics\\
Ben-Gurion University of the Negev\\
Beer-Sheva 84105 \\
Israel\\
\href{mailto:shevelev@bgu.ac.il}{\tt shevelev@bgu.ac.il} \\
\ \\
Charles R. Greathouse IV\\
3214 Whitethorn Road\\
Cleveland Heights, OH 44118\\
USA\\
\href{mailto:charles.greathouse@case.edu}{\tt charles.greathouse@case.edu}\\
\ \\
Peter J. C. Moses\\
Moparmatic Co.
1154 Evesham Road\\
Astwood Bank, Redditch, Worcestershire\\
B96 6DT England\\
\href{mailto:mows@mopar.freeserve.co.uk}{\tt mows@mopar.freeserve.co.uk}\\
\end{center}
\vskip .2 in
\begin{abstract}
We study values of $k$ for which the interval $(kn,(k+1)n)$ contains a
prime for every $n>1$. We prove that the list of such integers $k$
includes $ 1,2,3,5,9,14 $ and no others, at least for $k\leq
100,000,000$. Moreover, for every known $k$ in this list, we give a
good upper bound for the smallest $N_k(m),$ such that if $n\geq N_k(m),$
then the interval $(kn,(k+1)n)$ contains at least $m$ primes.
\end{abstract}
\section{Introduction and main results}
\label{intro}
In 1850, Chebyshev proved the famous Bertrand postulate (1845) that every interval $[n,2n]$ contains a prime (for a very elegant version of his proof, see Redmond \cite[Theorem 9.2]{10}). Other nice proofs were given by
Ramanujan in 1919 \cite{8} and Erd\H{o}s
in 1932 (reproduced in Erd\H{o}s and Sur\'{a}nyi \cite[p.\ 171--173]{4}). In 2006, El Bachraoui \cite{1} proved that every interval
$[2n,3n]$ contains a prime, while Loo \cite{6}
proved the same statement for every interval $[3n,4n]$. Moreover, Loo
found a lower bound for the number of primes in the interval $[3n,4n]$.
In 1952, Nagura \cite{7} proved that there is always a prime between $n$ and $\frac{6}{5}n$ for $n\geq25$. From his result, it follows that the interval $[5n,6n]$ always contains a prime. In this paper we prove the following:
\begin{theorem}\label{t1}
The list of integers \;$k$ for which every interval $(kn, (k+1)n),\; n>1,$ contains a prime includes\; $k=1,2,3,5,9,14$ and no others, at least for $k\leq 100,000,000$.
\end{theorem}
To prove Theorem \ref{t1}, in Section \ref{ramanujan} we introduce and study the so-called $k$-Chebyshev primes. We give them, and the generalized Ramanujan primes, the best estimates of the form $p_{tn},$ where $p_n$ is the $n$-th prime. Note that the core of the proof of Theorem \ref{t1} is Proposition \ref{p9}, which in turn depends on Proposition \ref{p8}.
In passing, for every $k=1,2,3,5,9,14,$ we give an algorithm for finding the smallest $N_k(m),$ such that for $n\geq N_k(m),$ the interval $(kn,(k+1)n)$ contains at least $m$ primes.
Proof of Theorem \ref{t1} is completed in Section \ref{proof} by computer research of sequence \seqnum{A218831} in \cite{13}.
\section{Case $k=1$}
\label{kone}
Ramanujan \cite{8} not only proved Bertrand's postulate, but also provided the smallest integers $\{R(m)\},$ such that if $x\geq R(m),$ then the interval $\left(\frac{x}{2}, x\right]$ contains at least $m$ primes, or equivalently, $\pi(x)-\pi(x/2)\geq m$. It is easy to see that it is sufficient to consider \emph{integer} $x$, and it is also evident that every term of $\{R(m)\}$ is prime. The numbers $\{R(m)\}$ are called \emph{Ramanujan primes} \cite{14}.
It is sequence \seqnum{A104272} in \cite{13}:
\begin{equation}\label{1}
2, 11, 17, 29, 41, 47, 59, 67, 71, 97,\ldots
\end{equation}
Since $\pi(x)-\pi(x/2)$ is not a monotonic function, to calculate the Ramanujan numbers one should have an effective upper bound for $R(m)$. Ramanujan \cite{8} showed that
\begin{equation}\label{2}
\pi(x)-\pi(x/2)>\frac{1}{\ln x}\left(\frac{x}{6}-3\sqrt{x}\right),\; x>300.
\end{equation}
In particular, for $x\geq324,$ the left-hand side is positive and thus $\geq1$. Using direct descent, he found that $\pi(x)-\pi(x/2)\geq1$ from $x\geq2$. Thus $R(1)=2$, which proves the Bertrand postulate. Further, e.g., for $x\geq400,$ the left-hand side of (\ref{2}) is more than 1 and thus $\geq2$. Again, using direct descent, he found that $\pi(x)-\pi(x/2)\geq2$ from $x\geq11$. Thus $R(2)=11,$ etc. Sondow \cite{14} found that $R(m)<4m\ln(4m)$ and conjectured that
\begin{equation}\label{3}
R(m)1$ be a real number. A $v$-Ramanujan number, $R_v(m),$ is the smallest integer such that if $x\geq R_v(m),$ then $\pi(x)-\pi(x/v)\geq m$.
\end{definition}
It is known \cite{10} that all $v$-Ramanujan numbers are primes. In particular, $R_2(m)=R(m),\; m=1,2,\ldots$ are the proper Ramanujan primes.
\begin{definition}\label{d4}
For a real number $v>1$ the \emph{$v$-Chebyshev number}, $C_v(m)$, is the smallest integer such that if $x\geq C_v(m),$ then $\vartheta(x)-\vartheta(x/v)\geq m\ln x,$ where $\vartheta(x)=\sum_{p\leq x}\ln p$ is the Chebyshev function.
\end{definition}
Since $\frac{\vartheta(x)-\vartheta(x/v)}{\ln x}$ can increase by 1 only when $x$ is prime, then all $v$-Chebyshev numbers are primes.
\begin{proposition}\label{p5}
We have
\begin{equation}\label{6}
R_v(m)\leq C_v(m).
\end{equation}
\end{proposition}
\begin{proof} Let $x\geq C_v(m)$. Then we have
\begin{equation}\label{7}
m\leq \frac {\vartheta(x)-\vartheta(x/v)}{\ln x}=\sum_{\frac{x}{v}

1, \;m\geq1$. In particular, if $k=1$ we find $\{C_2(m)\}$:
\begin{eqnarray}
&& 11,17,29,41,47,59,67,71,97,101,107,127,149,151,167,179,223, \nonumber \\
&& 229,233,239,241,263,269,281,307,311,347,349,367,373,401,409, \label{10} \\
&& 419,431,433,443,\ldots \nonumber
\end{eqnarray}
This sequence requires a separate comment. We observe that up to $C_2(100)=1489$ only two terms of this sequence $(C_2(17)=223$ and $C_2(36)=443)$ are not Ramanujan numbers, and the sequence is missing only the following 6 Ramanujan numbers: 181,227,439,491,1283,1301 and no others up to the 104-th Ramanujan number 1489. The latter observation shows how much the ratio $\frac{\vartheta(x)}{\ln x} $ exactly approximates $\pi(x)$. Similar observations are also valid for the following sequences for $v=\frac{k+1}{k}$ (and undoubtedly require an additional special study):
for $k=2,\;\{C_v(m)\},$
\begin{equation}\label{11}
13, 37, 41, 67, 73, 97, 127, 137, 173, 179, 181, 211, 229, 239, \ldots ;
\end{equation}
for $k=2,\;\{R_v(m)\},$
\begin{equation}\label{12}
2, 13, 37, 41, 67, 73, 97, 127, 137, 173, 179, 181, 211, 229, 239, \ldots ;
\end{equation}
for $k=3,\;\{C_v(m)\},$
\begin{equation}\label{13}
29, 59, 67, 101, 149, 157, 163, 191, 227, 269, 271, 307, 379, \ldots ;
\end{equation}
for $k=3,\;\{R_v(m)\},$
\begin{equation}\label{14}
11, 29, 59, 67, 101, 149, 157, 163, 191, 227, 269, 271, 307, 379, \ldots ;
\end{equation}
for $k=5,\;\{C_v(m)\},$
\begin{equation}\label{15}
59, 137, 139, 149, 223, 241, 347, 353, 383, 389, 563, 569, 593, \ldots ;
\end{equation}
for $k=5,\;\{R_v(m)\},$
\begin{equation}\label{16}
29, 59, 137, 139, 149, 223, 241, 347, 353, 383, 389, 563, 569, 593, \ldots ;
\end{equation}
for $k=9,\;\{C_v(m)\}$,
\begin{equation}\label{17}
223, 227, 269, 349, 359, 569, 587, 593, 739, 809, 857, 991, 1009, \ldots ;
\end{equation}
for $k=9,\;\{R_v(m)\},$
\begin{equation}\label{18}
127, 223, 227, 269, 349, 359, 569, 587, 593, 739, 809, 857, 991, 1009, \ldots ;
\end{equation}
for $k=14,\;\{C_v(m)\},$
\begin{equation}\label{19}
307, 347, 563, 569, 733, 821, 1427, 1429, 1433, 1439, 1447, 1481, \ldots ;
\end{equation}
for $k=14,\;\{R_v(m)\},$
\begin{equation}\label{20}
127, 307, 347, 563, 569, 733, 1423, 1427, 1429, 1433, 1439, 1447, \ldots
\end{equation}
\begin{remark}\label{r7}
In fact, Dusart \cite[Theorem 5.2]{3} gives several inequalities of the form
$$|\vartheta(x)-x|\leq\frac{ax}{\ln^bx},\; x\geq x_0(a,b)$$
From a computing point of view, the values $a=1300, \;b=4$ from Dusart's theorem are not always the best. The analysis for $x\geq25$ shows that the condition
$$x\left(1-\frac{1}{v}\right)\left(1-\frac{ax}{\ln^bx}\right)\geq m\ln x$$
is satisfied for the smallest $x_v(m)=x_v(m;a,b),$ using the following values of $a$ and $b$ from Dusart's theorem:
$a=3.965,\;b=2$ for $x$ in range $(25,\;7\cdot10^7];$
$a=\;1300,\;b=4$ for $x$ in range $(7\cdot10^7,\;10^9];$
$a=0.001,\;b=1$ for $x$ in range $(10^9,\;8\cdot10^9];$
$a=\;\;0.78,\;b=3$ for $x$ in range $(8\cdot10^9,\;7\cdot 10^{33}];$
$a=\;1300,\;b=4$ for $x>7\cdot 10^{33},$ \\
which minimizes the amount of calculations for $v$-Chebyshev primes.
\end{remark}
\section{Bounds of type (\ref{3})}
\label{bounds1}
\begin{proposition}\label{p8}
We have
\begin{equation}\label{21}
C_2(m-1)\leq p_{3m},\;m\geq2;
\end{equation}
\begin{equation}\label{22}
R_{\frac{3}{2}}(m)\leq p_{4m},\; m\geq1;\; C_{\frac{3}{2}}(m-1)\leq p_{4m},\; m\geq2;
\end{equation}
\begin{equation}\label{23}
R_{\frac{4}{3}}(m)\leq p_{6m},\; m\geq1;\;C_{\frac{4}{3}}(m-1)\leq p_{6m},\; m\geq2;
\end{equation}
\begin{equation}\label{24}
R_{\frac{6}{5}}(m)\leq p_{11m},\; m\geq1;\;C_{\frac{6}{5}}(m-1)\leq p_{11m},\; m\geq2;
\end{equation}
\begin{equation}\label{25}
R_{\frac{10}{9}}(m)\leq p_{31m},\; m\geq1;\;C_{\frac{10}{9}}(m-1)\leq p_{31m},\; m\geq2;
\end{equation}
\begin{equation}\label{26}
R_{\frac{15}{14}}(m)\leq p_{32m},\; m\geq1;\;C_{\frac{15}{14}}(m-1)\leq p_{32m},\; m\geq2.
\end{equation}
\end{proposition}
\begin{proof}
Firstly, let us find some values of $m_0=m_0(k),$ such that, at least, for $m\geq m_0$ all formulas (\ref{21})--(\ref{26}) hold. According to (\ref{8}) and (\ref{9}), it is sufficient to show that, for $m\geq m_0,$ we can take $p_{tm},$ where $t=3,4,6,11,31,32$ for formulas (\ref{21})-(\ref{26}) respectively, in the capacity of $x_v(m)$. As we noted in Remark \ref{r7}, in order to find possibly smaller values of $m_0,$ we use the bound
\begin{equation}\label{27}
\frac{x}{\ln x}\left(1-\frac{3.965}{\ln^2x}\right)\geq \frac {vm}{v-1}
\end{equation}
instead of (\ref{8}).
In order to get $x=p_{mt}$ satisfying this inequality, note that \cite{11}
$$ p_n\geq n\ln n.$$
Therefore, it is sufficient to consider $p_{mt}$ satisfying the inequality
$$\ln p_{tm}\leq\left(1-\frac{1}{v}\right)t\ln (tm)\left(1-\frac{3.965}{\ln^2(tm\ln (tm))}\right). $$
On the other hand, for $n\geq2,$ (see \cite[(4.2)]{3})
$$\ln p_n\leq\ln n+\ln\ln n+1. $$
Thus, it is sufficient to choose $m$ so large that the following inequality holds
$$\ln (tm)+\ln\ln (tm)+1\leq\left(1-\frac{1}{v}\right)t\ln (tm)\left(1-\frac{3.965} {\ln^2(tm\ln (tm))}\right),$$
or, since $1-\frac{1}{v}=\frac{1}{k+1},$ that
\begin{equation}\label{28}
\frac {\ln (tm)+\ln\ln (tm)+1}{\ln (tm)(1-\frac{3.965}{\ln^2(tm\ln (tm))})}\leq \frac{t}{k+1}.
\end{equation}
For example, let $k=1,\; t=3$. We can choose $m_0=350$. Then the left-hand side of (\ref{28}) equals $1.4976\cdots<1.5 $. This means that at least for $m\geq350,$ the estimate (\ref{21}) is valid. Using a computer for $m\leq350,$ we obtain (\ref{21}) for $m\geq2$. Other bounds are proved in the same way.
\end{proof}
\section{Bounds and formulas for $N_k(m)$}
\label{bounds2}
\begin{proposition}\label{p9}
\begin{equation}\label{29}
N_k(1)=2,\; k=2,3,5,9,14.
\end{equation}
For $m\geq2,\; k\geq1,$
\begin{equation}\label{30}
N_k(m)\leq\left\lceil\frac{R_{\frac{k+1}{k}}(m)}{k+1}\right\rceil;
\end{equation}
besides, if $R_{\frac{k+1}{k}}(m)\equiv1\pmod{k+1},$ then
\begin{equation}\label{31}
N_k(m)=\left\lceil\frac{R_{\frac{k+1}{k}}(m)}{k+1}\right\rceil=\frac{R_{\frac{k+1}{k}}(m)+k}{k+1}
\end{equation}
and, if $R_{\frac{k+1}{k}}(m)\equiv2\pmod{k+1},$ then
\begin{equation}\label{32}
N_k(m)=\left\lceil\frac{R_{\frac{k+1}{k}}(m)}{k+1}\right\rceil=\frac{R_{\frac{k+1}{k}}(m)+k-1}{k+1}.
\end{equation}
\end{proposition}
\begin{proof}
If $m\geq2,$ formally, the condition $x=(k+1)n\geq (k+1)N_k(m)$ is not stronger than the condition $x\geq R_{\frac{k+1}{k}}(m),$ since the first one is valid only for $x$ multiple of $k+1$. Therefore, for $m\geq2,$ (\ref{30}) holds. It allows calculation of terms in the sequence $\{N_k(m)\}$ for $k>1, \;m\geq2$. Since $N_k(1)\leq N_k(2),$ then, having $N_k(2),$ we can also prove (\ref{29}) using direct calculation. Now let $R_{\frac{k+1}{k}}(m)\equiv1\pmod{k+1}$. Note, that for $y=(R_{\frac{k+1}{k}}(m)-1)/(k+1)$ the interval
\begin{equation}\label{33}
(ky,\;(k+1)y)=\left(\frac{k}{k+1}\left(R_{\frac{k+1}{k}}(m)-1\right),\; R_{\frac{k+1}{k}}(m)-1\right)
\end{equation}
cannot contain more than $m-1$ primes. Indeed, it is an interval of type $\left(\frac{k}{k+1}x,x\right)$ for integer $x$, and the following such interval is
$$\left(\frac{k}{k+1}\left(R_{\frac{k+1}{k}}(m)\right), R_{\frac{k+1}{k}}(m)\right).$$
By definition, $R_{\frac{k+1}{k}}(m)$ is the \emph{smallest} number such that if $x\geq R_{\frac{k+1}{k}}(m),$ then $\{(\frac{k}{k+1}x, x)\}$ contains $\geq m$ primes. Therefore, the supposition that the interval (\ref{33}) contains $\geq m$ primes contradicts the minimality of $R_{\frac{k+1}{k}}(m)$. Since the following interval of type $(ky,\;(k+1)y)$ with integer $y\geq \frac{k}{k+1}(R_{\frac{k+1}{k}}(m)-1)$ is
$$\left(\frac{k}{k+1}(R_{\frac{k+1}{k}}(m)+k),\;R_{\frac{k+1}{k}}(m)+k\right),$$
then (\ref{31}) follows.
Finally, let $R_{\frac{k+1}{k}}(m)\equiv2\pmod{k+1}$. Again, for $y=(R_{\frac{k+1}{k}}(m)-2)/(k+1)$ the interval
\begin{equation}\label{34}
(ky,\;(k+1)y)=\left(\frac{k}{k+1}(R_{\frac{k+1}{k}}(m)-2),\;R_{\frac{k+1}{k}}(m)-2\right)
\end{equation}
cannot contain more than $m-1$ primes. Indeed, comparing interval (\ref{34}) with interval (\ref{33}), we see that they contain the same integers except for $R_{\frac{k+1}{k}}(m)-2$, which is multiple of $k+1$. Therefore, they contain the same number of primes, and this number does not exceed $m-1$. Again, since the following interval of type $(ky,\;(k+1)y)$ with integer $y\geq \frac{k}{k+1}(R_{\frac{k+1}{k}}(m)-2)$ is
$$\left(\frac{k}{k+1}(R_{\frac{k+1}{k}}(m)+k-1),\;R_{\frac{k+1}{k}}(m)+k-1\right),$$
then (\ref{32}) follows.
\end{proof}
As a corollary from (\ref{29}), (\ref{31}) and (\ref{32}), we obtain the following formula in case $k=2$.
\begin{proposition}\label{p10}
\begin{equation}\label{35}
N_2(m)=\begin{cases}
2, & \text{if $m=1$};\\
\left\lceil\frac{R_{\frac{3}{2}}(m)}{3}\right\rceil, & \text{if $ m\geq 2$.}
\end{cases}
\end{equation}
\end{proposition}
Note that, if $k\geq3$ and $R_{\frac{k+1}{k}}(m)\equiv j\pmod{k+1},\; 3\leq j\leq k,$ then, generally speaking, (\ref{30}) is not an equality. Evidently, $N_k(m)\geq N_k(m-1)$, and it is interesting that the equality is attainable
(see sequences (\ref{37})--(\ref{40}) below).
\begin{example}\label{e11} Let $k=3,\; m=2$. Then $v=\frac{4}{3}$ and, by (\ref{14}), $R_{\frac{4}{3}}(2)=29\equiv1\pmod4$. Therefore, by (\ref{31}),
$N_3(2)=\frac{29+3}{4}=8$. Indeed, interval $(3\cdot7,\; 4\cdot7)$ already contains only prime 23.
\end{example}
\begin{example}\label{e12} Let $k=3,\; m=3$. Then, by (\ref{14}), $R_{\frac{4}{3}}(3)=59\equiv3\pmod4$. Here $N_3(3)=11$ which is essentially less than
$\left\lceil {R_{\frac{4}{3}}(3)/4}\right\rceil=15$. Indeed, each interval
$$(3\cdot15,\; 4\cdot15),\;(3\cdot14,\; 4\cdot14),\;(3\cdot13,\; 4\cdot13),\;(3\cdot12,\; 4\cdot12),\; (3\cdot11,\; 4\cdot11)$$
contains more than 2 primes and only interval $(3\cdot10,\; 4\cdot10)$ contains only 2 primes.
\end{example}
In any case, Proposition~\ref{p9} allows us to calculate terms of sequence $\{N_k(m)\}$ for every considered value of $k$. So, we obtain the following few terms of $\{N_k(m)\}:$
for $k=2,$
\begin{equation}\label{36}
2,5,13,14,23,25,33,43,46,58,60,61,71,77,80,88,103,104, \ldots ;
\end{equation}
for $k=3,$
\begin{equation}\label{37}
2,8,11,17,26,38,40,41,48,57,68,68,70,87,96,100,108,109, \ldots ;
\end{equation}
for $k=5,$
\begin{equation}\label{38}
2,7,17,24,25,38,41,58,59,64,65,73,95,97,103,106,107,108, \ldots ;
\end{equation}
for $k=9,$
\begin{equation}\label{39}
2,14,23,23,34,36,57,58,60,60,77,86,100,100,102,123,149, \ldots ;
\end{equation}
for $k=14,$
\begin{equation}\label{40}
2,11,24,37,38,39,50,96,96,96,96,97,97,125,125,132,178,178, \ldots
\end{equation}
\begin{remark}\label{r13}
If, as in \cite{1,6}, instead of intervals $(kn,\;(k+1)n),$ we consider intervals $[kn,\;(k+1)n],$ then sequences (\ref{5}) and (\ref{36})--(\ref{38}) would begin with 1.
\end{remark}
\section{Method of small intervals}
\label{small}
If we know a theorem of the type: for $x\geq x_0(\Delta),$ the interval $(x,\; (1+\frac{1}{\Delta})x]$ contains a prime, then we can calculate a bounded number of the first terms of sequences (\ref{5}) and (\ref{36})--(\ref{40}). Indeed, put $x_1=kn,$ such that $n\geq\frac{x_0}{k}$. Then $(k+1)n=\frac{k+1}{k}x_1$ and, if $1+\frac{1}{\Delta}k,$ then
$$\left(x_1, \;(1+\frac{1}{\Delta})x_1\right]\subset(kn, (k+1)n). $$
Thus, if $n\geq\frac{x_0}{k},$ then the interval $(kn, (k+1)n)$ contains a prime, and using method of finite descent, we can find $N_k(1)$. Further, put $x_2=(1+\frac{1}{\Delta})x_1$. Then interval $(x_2,\; (1+\frac{1}{\Delta})x_2]$ also contains a prime. Thus the union
$$\left(x_1,\;(1+\frac{1}{\Delta})x_1\right]\cup\left(x_2,\;(1+\frac{1}{\Delta})x_2\right]=\left(x_1,(1+\frac{1}{\Delta})^2x_1\right]$$
contains at least two primes. This means that if
$(1+\frac{1}{\Delta})^2x_110726905041,$ or, equivalently, the interval $(x,\; (1+28313999^{-1})x)$ contains a prime if $x>10726905419$. This means that, e.g., we can find $N_{14}(m)$ for $m \le 1954471$. Unfortunately, this method cannot give the exact bounds and formulas for $N_k(m)$ as (\ref{30})--(\ref{32}).
We can also consider a more general application of this method. Consider a fixed infinite set $P$ of primes which we call $P$-primes. Furthermore, consider the following generalization of $v$-Ramanujan numbers.
\begin{definition}\label{d14}
For $v>1,$ a $(v,P)$-Ramanujan number, $R^{(P)}_v(m),$ is the smallest integer such that if $x\geq R^{(P)}_v(m),$ then $\pi_P(x)-\pi_P(x/v)\geq m,$ where $\pi_P(x)$ is the number of $P$-primes not exceeding $x$.
\end{definition}
Note that every $(v,P)$-Ramanujan number is $P$-prime. If we know a theorem of the type: for $x\geq x_0(\Delta),$ the interval
$\left(x,\; (1+\frac{1}{\Delta})x\right]$ contains a $P$-prime, then using the above described algorithm, we can calculate a bounded number of the first $(v,P)$ -Ramanujan numbers. For example, let $P$ be the set of primes $p\equiv1\pmod{3}$. From the result of Cullinan and Hajir \cite{2} it follows, in particular, that for $x\geq106706,$ the interval $(x,\; 1.048x)$ contains a $P$-prime. Using the same algorithm, we can calculate the first 14 $(2,P)$-Ramanujan numbers. They are
\begin{equation}\label{41}
7,31,43,67,97,103,151,163,181,223,229,271,331,337.
\end{equation}
Analogously, if $P$ is the set of primes $p\equiv2\pmod{3},$ then the sequence of $(2,P)$-Ramanujan numbers begins
\begin{equation}\label{42}
11, 23, 47, 59, 83, 107, 131, 167, 227, 233, 239, 251, 263, 281, \ldots ;
\end{equation}
if $P$ is the set of primes $p\equiv1\pmod{4},$ then the sequence of $(2,P)$-Ramanujan numbers begins
\begin{equation}\label{43}
13, 37, 41, 89, 97, 109, 149, 229, 233, 241, 257, 277, 281, 317, \ldots ;
\end{equation}
and, if $P$ is the set of primes $p\equiv3\pmod{4},$ then the sequence of $(2,P)$-Ramanujan numbers begins
\begin{equation}\label{44}
7, 23, 47, 67, 71, 103, 127, 167, 179, 191, 223, 227, 263, 307, \ldots ;
\end{equation}
Let $N^{(P)}_k(m)$ denote the smallest number such that for $n\geq N^{(P)}_k(m),$ the interval $(kn,(k+1)n)$ contains at least $m$ $P$-primes. It is easy to see that formulas (\ref{30})--(\ref{32}) hold for $N^{(P)}_k(m)$ and $R^{(P)}_{\frac{k+1}{k}}(m)$. In particular, in cases $k=1,2$ we have the formulas
\begin{equation}\label{45}
N^{(P)}_1(m)=\frac{R^{(P)}_2(m)+1}{2},\;\;N^{(P)}_2(m)=\left\lceil\frac{R^{(P)}_{\frac{3}{2}}(m)}{3}\right\rceil.
\end{equation}
Therefore, the following sequences for $N^{(P)}_1(m)$, correspond to
sequences (\ref{41})--(\ref{44}) respectively:
\begin{equation}\label{46}
4,16,22,34,49,52,76,82,91,112,115,136,166, 169, \ldots ;
\end{equation}
\begin{equation}\label{47}
6,12,24,30,42,54,66,84,114,117,120,126,132,141, \ldots ;
\end{equation}
\begin{equation}\label{48}
7,19,21,45,49,55,75,115,117,121,129,139,141,159, \ldots ;
\end{equation}
\begin{equation}\label{49}
4,12,24,34,36,52,64,84,90,96,112,114,132,154, \ldots
\end{equation}
\section{The proof of Theorem \ref{t1}}
\label{proof}
For $k\geq1,$ let $a(k)$ denote the least integer $n>1$ for which the interval $(kn,\;(k+1)n)$ contains no prime; if no such $n$ exists, we put $a(k)=0$. Taking into account (\ref{29}), note that $a(k)=0$ for $k=1,2,3,5,9,14$. Consider sequence $\{a(k)\}$. Its first few terms are (\seqnum{A218831} in \cite{13})
\begin{equation}\label{50}
0,0,0,2,0,4,2,3,0,2,3,2,2,0,6,2,2,3,2,6,3,2,4,2,2,7,2,2,4,3, \ldots
\end{equation}
Calculations of $a(k)$ for $k\in [1,15],$ except for $k=1,2,3,5,9,14$, give positive values of $a(k)$. Computer calculations of $a(k)$ in the range $\{16,\ldots ,10^8\}$ show that all values of $a(k)$ in this range are positive and belong to the interval $[2,16]$. This completes the proof.\qed
In conclusion, we present a distribution of numbers of values $a(k)=2,3,\ldots ,16$ within intervals $\{[1, 10^7(i)]\}, i=1,\ldots ,10$. All these numerical results are obtained using the following \emph{Mathematica} program:
\begin{lstlisting}[language=Mathematica]
start=2;cutOff=100;
a218831=Table[
NestWhile[#+1&,start,
Union[PrimeQ[Range[# k+1,# (k+1)-1]]]!={False}&,
1,cutOff],
{k,1,100000000}]/.{cutOff+start->0};
\end{lstlisting}
We have for $a(k)=2$ the numbers
\begin{eqnarray*}
&& 8729394,17566347,26437886,35330619,44238546, \\
&& 53158353,62087802,71025543,79969616,88921064.
\end{eqnarray*}
In general, here we have a simple explicit formula: the number of $a(k)=2, k\leq K$ is $K+1-\pi(2K+1)$. Further, let
$$ A_t(K)=|\{k\leq K: a(k)=t\}|.$$
In cases $t\geq3$ we have no explicit formulas. But, taking into account the distribution of primes into residue classes, a rough argument suggests that $A_t(K)\asymp c_t K(\ln K)^{2-t}$. For example, for $a(k)=3$ within the considered intervals we have the numbers
\begin{eqnarray*}
&& 1061880,2050703,3014798,3963752,4901317, \\
&& 5830488,6752801,7668802,8580597,9486975,
\end{eqnarray*}
and one can hope that $c_3\approx 1.7\cdots $. In other cases we have
$$ t=4:\;173835,321315,461745,597249,729660,859605,987238,1113288,1237558,1360344;$$
$$ t=5:\;25108,45086,63177,80407,97199,113213,128850,144474,159648,174577;$$
$$ t=6:\;7312,12542,17150,21536,25714,29734,33616,37243,40952,44503;$$
$$ t=7:\;1753,2918,3841,4749,5590,6373,7201,7950,8691,9378;$$
$$ t=8:\;449,703,918,1109,1309,1507,1670,1810,1977,2141;$$
$$ t=9:\;149,216,278,342,400,440,508,558,606,647;$$
$$ t=10:\;73,109,138,164,186,203,222,232,249,262;$$
$$ t=11:\;18,25,29,31,35,36,42,46,48,49;$$
$$ t=12:\;13,15,17,19,21,25,26,29,30,31;$$
$$ t=13:\;2,3,3,3,3,3,3,4,7,7;$$
$$ t=14:\;4,5,6,6,6,6,7,7,7,8;$$
$$ t=15:\;0,2,3,3,3,3,3,3,3,3;$$
$$ t=16:\;4,5,5,5,5,5,5,5,5,5.$$
For those $t$ when the difference
$$\frac{A_t(10^8)}{10^8}(8\ln10)^{t-2}-\frac{A_t(10^7)}{10^7}(7\ln10)^{t-2}$$
remains less than 0.5, we can get an impression about the change of $c_t$ depending on $t$. So, $c_2=1$ and approximately $c_3=1.7, \; c_4=4.6,\; c_5=11, \;c_6=49$.
\section{Acknowledgment}
The authors are grateful to N. J. A. Sloane for several important remarks. They
thank also an anonymous referee for useful and constructive comments
that promoted an essential improvement of the paper.
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\end{thebibliography}
\bigskip
\hrule
\bigskip
\noindent 2010 {\it Mathematics Subject Classification}: Primary 11A41.
\noindent \emph{Keywords: } prime number, generalized Ramanujan prime.
\bigskip
\hrule
\bigskip
\noindent (Concerned with sequences
\seqnum{A084140},
\seqnum{A104272},
\seqnum{A164952},
\seqnum{A185004},
\seqnum{A185005},
\seqnum{A185006},
\seqnum{A185007},
\seqnum{A218769},
\seqnum{A218831},
\seqnum{A218850},
\seqnum{A220268},
\seqnum{A220269},
\seqnum{A220273},
\seqnum{A220274},
\seqnum{A220281},
\seqnum{A220293},
\seqnum{A220462},
\seqnum{A220463},
\seqnum{A220474}, and
\seqnum{A220475}.)
\bigskip
\hrule
\vspace*{+.1in}
\noindent
Received January 1 2013;
revised versions received May 7 2013; June 10 2013; July 25 2013.
Published in {\it Journal of Integer Sequences}, July 31 2013.
\bigskip
\hrule
\bigskip
\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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\end{document}